Question Number 180786 by Vynho last updated on 17/Nov/22
$${f}\left({t}\right)=\int_{\mathrm{0}} ^{{t}} {x}−\lfloor{x}\rfloor\:\:{dx} \\ $$
Answered by mr W last updated on 17/Nov/22
$${let}\:{n}=\lfloor{t}\rfloor \\ $$$${f}\left({t}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\int_{{k}} ^{{k}+\mathrm{1}} \left({x}−\lfloor{x}\rfloor\right){dx}+\int_{{n}} ^{{t}} \left({x}−\lfloor{x}\rfloor\right){dx} \\ $$$${f}\left({t}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \left({k}+\xi−{k}\right){d}\left({k}+\xi\right)+\int_{\mathrm{0}} ^{{t}−{n}} \left({n}+\xi−{n}\right){d}\left({n}+\xi\right) \\ $$$${f}\left({t}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \xi{d}\xi+\int_{\mathrm{0}} ^{{t}−{n}} \xi{d}\xi \\ $$$${f}\left({t}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\left({t}−{n}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$${f}\left({t}\right)=\frac{{n}}{\mathrm{2}}+\frac{\left({t}−{n}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$${f}\left({t}\right)=\frac{\lfloor{t}\rfloor+\left({t}−\lfloor{t}\rfloor\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$